design of Reinforce concrete Torsion in Beams Slides

8/10/2019 design of Reinforce concrete Torsion in Beams Slides

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CHAPTER 2

TORSION IN RC BEAMS

Prepared By Mesfin D.

1


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Torsion when encountered in reinforced concrete members

usually occurs in combination with flexure and transverseshear. Due to this Torsion in its pure form (generally

associated with metal shafts) is rarely encountered in

reinforced concrete.

Equi libr ium tors ion and compatib i li ty tors ion This is associated with twisting moments that are developed

in a structural member to maintain static equilibrium with the

external loads.

are independent of the torsional stiffness of the member. Such torsion must be necessarily considered in design.

INTRODUCTION


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Such type of torsion is induced in beams supporting lateral

overhanging projections, in beams curved in plansubjected to gravity loads, and in beams where the

transverse loads are eccentric with respect to the shear

centre of the cross-section.

fig.1a). Example of

Equilibrium torision


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This is the name given to the type of torsion induced in a

structural member by rotations (twists) applied at one ormore points along the length of the member.

The twisting moments induced are directly dependent on

the torsional stiffness of the member.

These moments are generally statically indeterminate.

Their analysis necessarily involves (rotational)

compatibility conditions; hence the name compatibility

torsion.

Torsions due to compatibility torsion are not necessaryfor equilibrium and may be neglected in ultimate limit

state calculations and Design.

Compatib i li ty Tors ion

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Members subjected to a torsional moment develop shear

stresses.

In general, these tend to increase in magnitude from the

longitudinal axis of the member to its surface.

If the shear stresses are sufficiently large, cracks will

propagate through the member and, if torsion

reinforcement is not provided, the member will collapsesuddenly.

circu lar member sub jected to a torque T

Torsional s tresses

6

a). Circular member b). Member in Torsion c . Shear distribution


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shear stress at a given cross-section varies linearly from the

longitudinal axis of the member to a maximum value, max, atthe periphery of the section.

The stress at any distance r from the longitudinal axis of a

circular member is given by:

The maximum shear stress, maxis found by setting r= /2 in

equation (1).

7

34max

16

32/

2

TT


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But for non-circular member, the distribution of shearstress is not straightforward.

rectangular member sub jected to a torque T

Contd

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Contd

9

Unlike in the circular member, the stress distribution in arectangular member is non-linear.

The maximum shear stress occurs at the mid-point of the

longer side.

Shear stress at the corners of the section is zero indicatingthat the corners of the section are not distorted under torsion.

Analytical studies have shown that the maximum shear stress,

max, in rectangular section is given by:

The parameter depends on the relative values of x and y.


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The stress distribution in thin-walled hollow members is

much easier to determine than for solid non-circularmembers.

The shear stress in the walls is reasonably constant and is

given by:

Thin-wal led ho l low member sub jected to a

torque T

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Where t is the thickness of the wall of the member and

Aois the area within a perimeter bounded by the centreline of the wall.

On a given section, the shear stress is maximum where

the thickness of the wall is minimum.

Fai lure of conc rete members w i th tors ion Consider a rectangular member subjected to a torque T

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Consider small elements on each face of the member, as

illustrated in the figure.As for members with applied shear, shear stresses act on

the sides of each element in the directions.

The equivalent principal stresses are inclined at an angle

of 450to the horizontal. In the same way as for shear, the principal tensile stresses

cause the development of cracks inclined at an angle of

450.

However, in the case of torsion, they form a spiral crack allaround the member

Since the shear stresses in members with torsion are

greatest at the surface, these cracks develop inwards from

the surface of the member.

(Contd...)

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For members with no form of reinforcement to preventthe opening of Torsional cracks, failure of the member

will occur almost as soon as the cracking begins.

Therefore, torsional failure of a member without

reinforcement is prevented only if the shear strength of

the concrete exceeds the shear stress due to applied

torsion.

The torsional strength of a concrete member can be

significantly increased by providing suitable torsion

reinforcement across the cracks.

This is usually provided in the form of closedfour-sided

stirrups, in combination with longitudinal bars distributed

around the periphery of the section.

(Contd...)

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This reinforcement controls the propagation of cracks and

ensures that when failure occurs due to yielding of the

reinforcement, it is not sudden. To quantify the behavior of members with such torsional

reinforcement, an equivalent space truss model, similar to

the plane truss model for shear, can be used.

(Contd...)

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This theory, developed by Lampert and Collins(1972),

assumes that solid members can be designed as equivalentthin walled hollow members .

The thickness of the wall, t, is commonly taken as:

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The concrete cracks are inclined at an angle which,

like shear, varies in the range 220< < 680.

The truss dimensions,xoand yo, are measured from

centre to centre of the notional thin walls.

(Contd...)

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Since the horizontal projection of the crack is yo(cot ),

the number of stirrups crossing the crack is yo(cot )/swhere sis the stirrup spacing.

Hence, assuming that reinforcement has yielded, the

total vertical shear force transmitted across the cracks by

the stirrups on one side face is:

Similarly the shear force transferred across the top or

bottom face is:

(Contd...)

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The shear force on the side wall of a thin walled member is

Substituting for from equation (3) to equation (7) gives:

Similarly it can be shown that:

(Contd...)

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Substituting from equation (5) into equation (9)

Thus the stirrup reinforcement required to resist torsion

of Tis:

In addition to the stirrup reinforcement, longitudinal

reinforcement is required to resist torsion.

(Contd...)

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From equilibrium of forces, the compressive force in the

diagonal members, C, must equal V1/sin.

And the force in the longitudinal members is

N = Ccos = V1cot.

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Substituting for V1

The total force in the longitudinal members from all four

joints at a given cross section is:

21

cot2cot2cot2 2121 VVVVNTotal

cot2 21 VVf

ANTotals

y

long

cotcot22

200

00

00 yx

yxTy

T

x

TfAs

y

long

cot

00

00

sy

long

fyx

yxTA


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Alternatively, it can be rearranged to give the maximum

torsion carried without leading to yielding of thelongitudinal reinforcement:

Regardless of how much stirrup and longitudinal

reinforcement is provided, the torsion must not be of

such magnitude as to cause crushing of the concrete in

the diagonal struts.

This force is resisted by stresses in the concrete

between the diagonal cracks.

22

cot00

00

yx

fyxAT

sy

long


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23

The surface area of concrete to which this is applied is,

from Fig. yo(cos ) t, Hence the stress in the struts is:

This must not exceed the compressive strength of

concrete, fck/c, where is the effectiveness factor for

torsion

cossincos

sin

0

1

0

1

ty

V

ty

V

strutinstress


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Substitution for V1 from equation (7) gives:

From which the torsion which could cause crushing of

the concrete struts is Tw, is:

24

cckf

tyV

cossin0

1

cckf

tyxT

cossin2

0

0

c

ck

w

ftyxT

cossin200

Members with combined actions


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For members with combined M and T, the longitudinal

reinforcement required for torsion should be provided in

addition to the amount required for moment.

For members with combined T and V, the ultimate

torque, T, and the ultimate shear force, V, should satisfy

the condition:

Members with combined actions

25

1

22

ww VV

TT

Design o f members for tors ion


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26

Design o f members for tors ion

in acco rdance w ith EBCS2


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The torsional resistance of members is calculated

on the basis of an equivalent thin-walled section (i.e.the truss analogy) as describe above.

Wall thickness t=teff=hef A/u the actual wall

thickness

For sections of complex shape, such as T-sections,

the torsional resistance can be calculated by

dividing the section into individual elements of

simple, say rectangular, shape.

The torsional resistance of the section is equal to

the sum of the capacities of the individual elements,

each modeled as an equivalent thin-walled section.

Contd.

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The critical section for torsion is at the face of supports.

Tsd does not exceed the torsional capacity, TRd whereTRd= 0.80fcdAefhef Aef= xoyo

The torsional resistance of concrete Tcshall be taken as:

The torsional resistance of the reinforcement Teffis given

by:

Members with pure torsion

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350mmor8

S

effU

Spacing

29

yk

wf

4.0min,

Members w i th combined


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Members subjected to combined moment and tors ion

The longitudinal reinforcement required for torsion andFlexure should be calculated separately and hence total

area of reinforcement to be provided should be the sum

of the two.

Members sub jected to combined shear and to rs ion

TRd and VRd(TRd= 0.80fcdAefhef and VRd= 0.25fcdbwd),

respectively, multiplied by the following reduction factors

tand v.

Members w i th combined

act ions

30

2

1

1

RdSd

RdSd

t

TT

VV

2

1

1

RdSd

RdSd

v

VV

TT

RdVvRdVRdTtRdT 'and'


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Tcand Vc(Tc= 1.2fctdAefhef and Vc= 0.25fctdk1k2bwd),

respectively, must be multiplied by the reduction factorstcand vc.

The calculations for the design of stirrups may be made

separately for torsion and shear and the total stirrup

required for shear and torsion are, of course, additive.

31

2

1

1

cSd

cSd

tc

TT

VV

2

1

1

cSd

cSd

vc

VV

TT

cVvcVTtcT ccc 'and'


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.

.

.

32

S

A

S

A

S

Atvtv

2

torsionandshearforstirrupleg2Total

ydeff

effst

fA

T

S

A

*2

)(

leg1 torsionforStirrup

df

V

S

A

yd

sdv

*

hear2 SforStirrupleg


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design of Reinforce concrete Torsion in Beams Slides